Question

2 point charges of +16 microcoulomb and -9 microcoulomb are placed 8 cm apart in air. Determine the position of the point where resultant electric field is 0 ( zero).

Answer 1

any comments ....wlcm

Answer 2

Answer:

The resultant electric field is zero at 24 cm.

Explanation:

Given that,

First point charge $q_{1} = 16\times10^{-6}\ C$

Second point charge $q_{2}= -9\times10^{-6}\ C$

Distance d = 8 cm

Let us consider the electric field will be zero at p.

The electric field at P by q₂

$E_{1} = \dfrac{F}{q_{2}}$

$E_{1}=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q_{1}}{r^2}$

$E_{1}=\dfrac{9\times10^{9}\times9\times10^{-6}}{x^2}$....(I)

The electric field at P by q₁

$E_{2}=\dfrac{9\times10^{9}\times16\times10^{-6}}{(8+x)^2}$....(II)

The electric field will be zero

$E_{1}+E_{2}=0$

$\dfrac{9\times10^{9}\times9\times10^{-6}}{x^2}+\dfrac{9\times10^{9}\times16\times10^{-6}}{(8+x)^2}=0$

$\dfrac{9\times10^{-6}}{x^2}=\dfrac{16\times10^{-6}}{(8-x)^2}$

$\dfrac{3}{x}=\dfrac{4}{8+x}$

$24+3x=4x$

$x = 24\ cm$

Hence, The resultant electric field is zero at 24 cm.